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20.66 ZTRANS: Z-Transform Package

This package is an implementation of the Z-transform of a sequence. This is the discrete analogue of the Laplace Transform.

Authors: Wolfram Koepf and Lisa Temme.

20.66.1 Z-Transform

The Z-Transform of a sequence {fn} is the discrete analogue of the Laplace Transform, and

∑∞ 𝒵 {fn} = F (z) = fnz−n. n=0

This series converges in the region outside the circle |z| = |z0| = limsupn→∞n∘ ---- |fn|.

SYNTAX:  ztrans(fn, n, z)        where fn is an expression, and n,z
are identifiers.

20.66.2 Inverse Z-Transform

The calculation of the Laurent coefficients of a regular function results in the following inverse formula for the Z-Transform:
If F(z) is a regular function in the region |z| > ρ then ∃ a sequence {fn} with 𝒵{fn} = F(z) given by

∮ -1-- n−1 fn = 2πi F (z)z dz

SYNTAX:  invztrans(F(z), z, n)        where F(z) is an expression,
and z,n are identifiers.

20.66.3 Input for the Z-Transform

This package can compute the Z-Transforms of the following list of fn, and certain combinations thereof.

1 eαn --1-- (n+k)
1- n! --1- (2n)! --1--- (2n+1)!
sin(βn) --n!-- sin(αn + ϕ) eαn sin(βn)
cos(nβ!n) cos(αn + ϕ) eαn cos(βn)
sin(β(n+1)) n+1 sinh(αn + ϕ)cos(β(n+1)) n+1
cosh(αn + ϕ) (n+k ) m

Other Combinations

Linearity

𝒵{afn + bgn} = a𝒵{fn} + b𝒵{gn}

Multiplication by n

𝒵{nk ⋅ fn} = −zd dz( k−1 ) 𝒵 {n ⋅fn,n, z}

Multiplication by λn

𝒵{λn ⋅ fn} = F(z) λ

Shift Equation

𝒵{fn+k} = zk( ) k−∑1 − j F (z)− j=0fjz

Symbolic Sums

𝒵{ } ∑n fk k=0 = -z- z− 1 ⋅𝒵{fn}
𝒵{ } n+∑q k=p fk   combination of the above

where k,λ ∈N∖{0}; and a,b are variables or fractions; and p,q ∈Z or are functions of n; and α,β and ϕ are angles in radians.

20.66.4 Input for the Inverse Z-Transform

This package can compute the Inverse Z-Transforms of any rational function, whose denominator can be factored over Q, in addition to the following list of F(z).

( ) (cos(β)) ( ) (cos(β) ) sin sinz(β)- e z cos sin(zβ) e z -- -- -- ∘ -zsin(∘ -z) cos(∘ z) A A A ∘ z- (∘ z) (∘ z-) A sinh A cosh A ( z ) (√z2+Az+B-) z log √z2−Az+B-- zlog ----z---- ( ) arctan zs+inc(oβs()β)

where k,λ ∈N ∖{0} and A,B are fractions or variables (B > 0) and α,β, and ϕ are angles in radians.

20.66.5 Application of the Z-Transform

Solution of difference equations

In the same way that a Laplace Transform can be used to solve differential equations, so Z-Transforms can be used to solve difference equations.
Given a linear difference equation of k-th order

fn+k + a1fn+k−1 + ...+ akfn = gn
(20.89)

with initial conditions f0 = h0, f1 = h1, …, fk−1 = hk−1 (where hj are given), it is possible to solve it in the following way. If the coefficients a1,…,ak are constants, then the Z-Transform of (20.89) can be calculated using the shift equation, and results in a solvable linear equation for 𝒵{fn}. Application of the Inverse Z-Transform then results in the solution of  (20.89).
If the coefficients a1,…,ak are polynomials in n then the Z-Transform of (20.89) constitutes a differential equation for 𝒵{fn}. If this differential equation can be solved then the Inverse Z-Transform once again yields the solution of (20.89). Some examples of these methods of solution can be found in §20.66.6.

20.66.6 EXAMPLES

Here are some examples for the Z-Transform

1: ztrans((-1)^n*n^2,n,z);  
 
    z*( - z + 1)  
---------------------  
  3      2  
 z  + 3*z  + 3*z + 1  
 
2: ztrans(cos(n*omega*t),n,z);  
 
   z*(cos(omega*t) - z)  
---------------------------  
                     2  
 2*cos(omega*t)*z - z  - 1  
 
3: ztrans(cos(b*(n+2))/(n+2),n,z);  
 
                                 z  
z*( - cos(b) + log(------------------------------)*z)  
                                          2  
                    sqrt( - 2*cos(b)*z + z  + 1)  
 
4: ztrans(n*cos(b*n)/factorial(n),n,z);  
 
  cos(b)/z       sin(b)                 sin(b)  
 e        *(cos(--------)*cos(b) - sin(--------)*sin(b))  
                   z                      z  
---------------------------------------------------------  
                            z  
5: ztrans(sum(1/factorial(k),k,0,n),n,z);  
 
  1/z  
 e   *z  
--------  
 z - 1  
 
6: operator f$  
 
7: ztrans((1+n)^2*f(n),n,z);  
 
                          2  
df(ztrans(f(n),n,z),z,2)*z  - df(ztrans(f(n),n,z),z)*z  
+ ztrans(f(n),n,z)

Here are some examples for the Inverse Z-Transform

8: invztrans((z^2-2*z)/(z^2-4*z+1),z,n);  
 
              n       n                n  
 (sqrt(3) - 2) *( - 1)  + (sqrt(3) + 2)  
-----------------------------------------  
                    2  
 
9: invztrans(z/((z-a)*(z-b)),z,n);  
 
  n    n  
 a  - b  
---------  
  a - b  
 
10: invztrans(z/((z-a)*(z-b)*(z-c)),z,n);  
 
  n      n      n      n      n      n  
 a *b - a *c - b *a + b *c + c *a - c *b  
-----------------------------------------  
  2      2        2      2    2        2  
 a *b - a *c - a*b  + a*c  + b *c - b*c  
 
11: invztrans(z*log(z/(z-a)),z,n);  
 
  n  
 a *a  
-------  
 n + 1  
 
12: invztrans(e^(1/(a*z)),z,n);  
 
        1  
-----------------  
  n  
 a *factorial(n)  
 
13: invztrans(z*(z-cosh(a))/(z^2-2*z*cosh(a)+1),z,n);  
 
cosh(a*n)

Examples: Solutions of Difference Equations

I
(See [BS81], p. 651, Example 1).
Consider the homogeneous linear difference equation
f − 2f + 2f − 3f + 2f = 0 n+5 n+3 n+2 n+1 n

with initial conditions f0 = 0, f1 = 0, f2 = 9, f3 = −2, f4 = 23. The Z-Transform of the left hand side can be written as F(z) = P(z)∕Q(z) where P(z) = 9z3 − 2z2 + 5z and Q(z) = z5 − 2z3 + 2z2 − 3z + 2 = (z − 1)2(z + 2)(z2 + 1), which can be inverted to give

fn = 2n + (− 2)n − cos π-n. 2

The following REDUCE session shows how the present package can be used to solve the above problem.

14: operator f$ f(0):=0$ f(1):=0$ f(2):=9$ f(3):=-2$ f(4):=23$  
 
 
20: equation:=ztrans(f(n+5)-2*f(n+3)+2*f(n+2)-3*f(n+1)+2*f(n),n,z);  
 
                              5                       3  
equation := ztrans(f(n),n,z)*z  - 2*ztrans(f(n),n,z)*z  
 
                                   2  
             + 2*ztrans(f(n),n,z)*z  - 3*ztrans(f(n),n,z)*z  
 
                                       3      2  
             + 2*ztrans(f(n),n,z) - 9*z  + 2*z  - 5*z  
 
 
21: ztransresult:=solve(equation,ztrans(f(n),n,z));  
 
                                             2  
                                       z*(9*z  - 2*z + 5)  
ztransresult := {ztrans(f(n),n,z)=----------------------------}  
                                    5      3      2  
                                   z  - 2*z  + 2*z  - 3*z + 2  
 
22: result:=invztrans(part(first(ztransresult),2),z,n);  
 
                   n    n       n    n  
           2*( - 2)  - i *( - 1)  - i  + 4*n  
result := -----------------------------------  
                          2

II
(See [BS81], p. 651, Example 2).
Consider the inhomogeneous difference equation:
fn+2 − 4fn+1 + 3fn = 1

with initial conditions f0 = 0, f1 = 1. Giving

F(z) = 𝒵{1}( 1 z ) z2-−-4z +-3 + z2 −-4z-+-3
= --z-- z − 1( ) ----1------+ ----z------ z2 − 4z + 3 z2 − 4z + 3.

The Inverse Z-Transform results in the solution

( n+1 ) fn = 1- 3---−--1− (n + 1) . 2 2

The following REDUCE session shows how the present package can be used to solve the above problem.

23: clear(f)$ operator f$ f(0):=0$ f(1):=1$  
 
 
27: equation:=ztrans(f(n+2)-4*f(n+1)+3*f(n)-1,n,z);  
 
                               3                       2  
equation := (ztrans(f(n),n,z)*z  - 5*ztrans(f(n),n,z)*z  
 
                                                           2  
    + 7*ztrans(f(n),n,z)*z - 3*ztrans(f(n),n,z) - z )/(z - 1)  
 
28: ztransresult:=solve(equation,ztrans(f(n),n,z));  
 
                                      2  
                                     z  
result := {ztrans(f(n),n,z)=---------------------}  
                              3      2  
                             z  - 5*z  + 7*z - 3  
 
29: result:=invztrans(part(first(ztransresult),2),z,n);  
 
              n  
           3*3  - 2*n - 3  
result := ----------------  
                 4

III
Consider the following difference equation, which has a differential equation for 𝒵{fn}.
(n+ 1)⋅fn+1 − fn = 0

with initial conditions f0 = 1, f1 = 1. It can be solved in REDUCE using the present package in the following way.

30: clear(f)$ operator f$ f(0):=1$ f(1):=1$  
 
 
34: equation:=ztrans((n+1)*f(n+1)-f(n),n,z);  
 
                                        2  
equation :=  - (df(ztrans(f(n),n,z),z)*z  + ztrans(f(n),n,z))  
 
35: operator tmp;  
 
36: equation:=sub(ztrans(f(n),n,z)=tmp(z),equation);  
 
                              2  
equation :=  - (df(tmp(z),z)*z  + tmp(z))  
 
37: load(odesolve);  
 
38: ztransresult:=odesolve(equation,tmp(z),z);  
 
                         1/z  
ztransresult := {tmp(z)=e   *arbconst(1)}  
 
39: preresult:=invztrans(part(first(ztransresult),2),z,n);  
 
              arbconst(1)  
preresult := --------------  
              factorial(n)  
 
40: solve({sub(n=0,preresult)=f(0),sub(n=1,preresult)=f(1)},  
arbconst(1));  
 
{arbconst(1)=1}  
 
41: result:=preresult where ws;  
 
                1  
result := --------------  
           factorial(n)

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